One leg of a right triangle is decreasing at a rate of $5$ kilometers per hour and the other leg of the triangle is increasing at a rate of $14$ kilometers per hour. At a certain instant, the decreasing leg is $3$ kilometers and the increasing leg is $9$ kilometers. What is the rate of change of the area of the right triangle at that instant (in square kilometers per hour)? Choose 1 answer: Choose 1 answer: (Choice A) A $111$ (Choice B) B $1.5$ (Choice C) C $-1.5$ (Choice D) D $-111$
Setting up the math Let... $l_1(t)$ denote the decreasing leg of the triangle at time $t$, $l_2(t)$ denote the increasing leg of the triangle at time $t$, and $A(t)$ denote the area of the triangle at time $t$. We are given that $l_1'(t)=-5$ and $l_2'(t)=14$ (notice that $l_1'$ is negative). We are also given that that $l_1(t_0)=3$ and $l_2(t_0)=9$ for a specific time $t_0$. We want to find $A'(t_0)$. Relating the measures The measures relate to each other through the formula for the area of a right triangle: $A(t)=\dfrac{l_1(t)l_2(t)}{2}$ We can differentiate both sides to find an expression for $A'(t)$ : $A'(t)=\dfrac{l_1'(t)l_2(t)+l_1(t)l_2'(t)}{2}$ Using the information to solve Let's plug ${l_1'(t_0)}={-5}$, ${l_2(t_0)}={9}$, ${l_1(t_0)}={3}$, and $C{l_2'(t_0)}=C{14}$ into the expression for $A'(t_0)$ : $\begin{aligned} A'(t_0)&=\dfrac{{l_1'(t_0)}{l_2(t_0)}+{l_1(t_0)}C{l_2'(t_0)}}{2} \\\\ &=\dfrac{({-5})({9})+({3})(C{14})}{2} \\\\ &=-1.5 \end{aligned}$ In conclusion, the rate of change of the area of the triangle at that instant is $-1.5$ square kilometers per hour. Since the rate of change is negative, we know that the area is decreasing.